Codebusters Test Codebusters Cipher Decryption 5

Question 1

The Hill cipher encrypts plaintext in blocks by multiplying them with a key matrix. For a 2×2 key matrix, how many plaintext letters are encrypted per block?

Accepted Answer

2

Answer

A 2×2 matrix multiplied by a 2×1 plaintext vector produces a 2×1 ciphertext vector, so two letters are encrypted at a time.

Question 2

In an affine cipher with encryption E(x) = (ax + b) mod 26, which condition must the multiplier 'a' satisfy for the cipher to be invertible?

Accepted Answer

gcd(a, 26) = 1

Answer

For a modular multiplicative inverse of 'a' to exist mod 26, 'a' and 26 must share no common factors.

Question 3

In RSA encryption with public key (e, n), how is a plaintext number m encrypted to produce ciphertext c?

Accepted Answer

c = m^e mod n

Answer

RSA encryption raises the plaintext to the public exponent e and reduces modulo n: c = m^e mod n.

Question 4

In RSA, the private key exponent d is computed so that ed ≡ 1 (mod φ(n)). Here φ(n) represents:

Accepted Answer

Euler's totient function of n

Answer

Euler's totient φ(n) counts integers up to n coprime to n; for n = pq it equals (p−1)(q−1).

Question 5

The Baconian cipher encodes each letter as a five-character sequence using only two symbols (A and B). How many distinct letters or symbols can a five-character binary sequence represent?

Accepted Answer

32

Answer

2^5 = 32 distinct sequences, which is more than enough to encode the 26 letters (or 24 when I/J and U/V are merged).

Codebusters Test Practice Test

Codebusters Test Practice Test

Codebusters Test Codebusters Cipher Decryption 5